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Well, not completely passive. But not what's usually meant when one uses the words 'active filter'.

It's going to be a filter that consists only of discrete transistors, capacitors and resistors.

The passband is going to be extremely narrow (50 Hz-1 kHz). And in LTspice I've found that when two poles are this close together the pass band becomes highly non-linear.

So my question is how do I design a filter where the response is very flat in the pass band?

Inductors could also be used if need be.

Ideally, it'll have at least 2 poles each for the LPF and HPF, but if this compromises linearity I'll just settle for the 1 pole I guess.

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    \$\begingroup\$ At first, you misunderstood something: non-linearity has nothing to do with pole location. Secondly, pleae make clear what you really want. A bandpass filter with which specification? Opamps allowed or transistors only? \$\endgroup\$
    – LvW
    Commented May 28, 2016 at 8:49
  • \$\begingroup\$ Well this is what I call a nice band pass response because of the flat plateau at the top. wa0itp.com/bpf_20m_plot.jpg If you see two poles close together in a passive filter though, I don't think it's possible for the plot to NOT look parabolic. It will generally look more like this mathsisfun.com/geometry/images/parabola-soccer.gif Also, weren't your other questions answered in my first post? \$\endgroup\$
    – gorge
    Commented May 28, 2016 at 8:59
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    \$\begingroup\$ I think, your specification is not clear (what is a "nice" response?). Each second order bandpass has no ripple in the passband. But you spoke about "at least" two poles. So - a fourth order bandpass could also meet your requirements? A bandpass specification requires (a) midband frequency, (b) midband gain, (c) Bandwidth, (d) damping requirements or transfer function order. \$\endgroup\$
    – LvW
    Commented May 28, 2016 at 9:13
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    \$\begingroup\$ Try specifying what you want with respect to this picture: images.books24x7.com/bookimages/id_14105/fig6-1.jpg \$\endgroup\$
    – Andy aka
    Commented May 28, 2016 at 10:11
  • \$\begingroup\$ Linearity is a property of the circuit itself. The use of a Bode plot as a means of expressing the transfer function of a circuit already presumes that the circuit is linear. Recall that a Bode plot exists and is unique for any given linear transfer function: it's an alternative means of expressing the transfer function. Nonlinear transfer functions cannot be represented with a Bode plot - it was never meant to be used that way. When it comes to filters, generally you want them very linear. Bode plots don't exist for nonlinear transfer functions. \$\endgroup\$ Commented Oct 29, 2021 at 13:50

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Well, not completely passive. But not what's usually meant when one uses the words 'active filter'.

You can't be "a little pregnant". A filter also can't be "not quite passive, but also not active": That's just nonsense.

It's going to be a filter that consists only of discrete transistors, capacitors and resistors.

Transistor in there, acting as transistor/diode and not as a passive component: Congratulations, this is not a passive circuit anymore. So let's discard this whole "passive" wording. It's simply wrong.

It's going to be a filter that consists only of discrete transistors, capacitors and resistors.

I'd call that a "filter made up of discrete transistors, capacitors and resistors". Nothing else.

The passband is going to be extremely narrow (50Hz-1kHz).

How is that extremely narrow? You're not specifying any transition widths or suppressions, but as this, you can implement it probably actually passively with a single- or dualstage LC bandpass sufficiently well.

And in LTspice I've found that when two poles are this close together the pass band becomes highly non-linear.

what, no! If you build a linear filter, it stays linear. If you use nonlinearities, then it really depends on your filter topology why and when it becomes nonlinear. Generally, poles don't have anything to do with this.

In fact, if you use poles to describe your filter, you might be doing something wrong if that filter is not linear – which you claim it would be – so really, I don't think your whole "using only discrete transistors" thing is clever at all. In fact, you can of course try to replicate a sufficiently well-working operational amplifier with transistors (it's hard because unlike on ICs, it's hard to find two closely equivalent discrete transistors), and actually build a useful filter.

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  • \$\begingroup\$ Active filter networks don't imply op-amps. A transistor can be a means of impedance scaling and power gain. Local feedback around the transistor will be stabilizing the transconductance. In well designed transistor-based filters, the passive elements' tolerances and thermal parameter drifts have more influence on the filter's response than the transistors' exact value of transconductance. But that's a bit of a truism: with few exceptions, circuits should accept a reasonable range of substitute parts and their associated parameter range. The circuit design should neutralize those. \$\endgroup\$ Commented Oct 29, 2021 at 13:56

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